Exercises — Tension in inextensible strings

Picture kya dikhati hai: ek block, ek string, kuch bhi nahi hil raha. Static kyun hai: "at rest" ka matlab hai acceleration $a=0$. Block ka FBD: do vertical forces — weight $mg$ neeche kheench raha hai, tension $T$ upar kheench rahi hai (string hamesha body ko apni taraf, yaani yahan upar kheenchti hai). $T-mg=ma=0\Rightarrow T=mg$ $T=3\times 10=\boxed{30\text{N}}$ Static shortcut "$T=mg$" sirf isliye theek hai kyunki $a=0$. Yeh yaad rakho — bilkul agli level mein yeh toot jaata hai.

Accelerations kyun match karni chahiye — step by step. Pulley se neeche latakti dono string lengths ko $x_1$ (block 1 tak) aur $x_2$ (block 2 tak) kaho. String stretch nahi ho sakti, toh total length fixed hai: $x_1+x_2=\text{constant}.$ Right-hand side time ke saath nahi badalti, toh uska rate of change zero hai. Ek baar time ke saath differentiate karo (position ke change ki rate = velocity): ${\dot{x}}_1+{\dot{x}}_2=0\Rightarrow {\dot{x}}_1=-{\dot{x}}_2(\text{ek end utna hi deta hai jitna doosra leta hai}).$ Dobara differentiate karo (velocity ke change ki rate = acceleration): ${\ddot{x}}_1=-{\ddot{x}}_2.$ Ab symbol $a$ ko in dono accelerations ki common magnitude define karo — ek single positive number jo dono blocks share karte hain. Symbols mein, $a\equiv |{\ddot{x}}_1|=|{\ddot{x}}_2|$, yaani $|a_1|=|a_2|=a.$ Upar ka minus sign geometry hai: ek neeche jaata hai jab doosra upar jaata hai, lekin magnitudes lock equal hain, aur $a$ us shared magnitude ka naam hai jo hum dete hain.

• (a) Galat — weights masses par depend karti hain, jo alag hain.
• (b) Sahi — $|a_1|=|a_2|=a$.
• (c) Galat — pulley ke upar ek neeche jaata hai jab doosra upar jaata hai; directions opposite hain. Answer: (b).

Yeh canonical Atwood Machine hai. Figure mein red block $m_1$ hai (baayi taraf ka bhaari wala), aur tum uska neeche ka motion arrow dekh sakte ho jabki white $m_2$ ka arrow upar hai — picture tum par yeh clearly bol rahi hai ki har body ke liye "positive" kahan point karein. Formula blindly quote karne ki jagah, do FBDs se isko rebuild karte hain taaki answer ki shape samajh aaye. $m_1$ ka FBD (bhaari red block, girta hai; positive = down): weight $m_{1}g$ neeche, tension $T$ upar. $m_{1}g-T=m_{1}a$ $m_2$ ka FBD (halka white block, utha jaata hai; positive = up): tension $T$ upar, weight $m_{2}g$ neeche. $T-m_{2}g=m_{2}a$ Inhe add kyun karein? Add karne se unknown $T$ cancel ho jaata hai aur sirf $a$ bachta hai: $m_{1}g-m_{2}g=(m_1+m_2)a\Rightarrow a=\frac{(m_1-m_2)g}{m_1+m_2}.$ Numerator net driving weight $(m_1-m_2)g$ hai; denominator total mass being moved hai — isliye formula (unbalanced force)/(total mass) jaisa dikhta hai, jo ki bas $F=ma$ hai. $a=\frac{(5-3)\times 10}{5+3}=\frac{20}{8}=\boxed{2.5{\text{m/s}}^2}$ Back-substitute karo $T-m_{2}g=m_{2}a$ mein tension ke liye: $T=m_2(g+a)=3(10+2.5)=\boxed{37.5\text{N}}$ (Equivalently $T=\frac{2m_1{m}_2}{m_1+m_2}g=\frac{2\cdot 5\cdot 3}{8}\times 10=37.5\text{N}$.) Sanity check: $T=37.5\text{N}$ $m_{2}g=30\text{N}$ aur $m_{1}g=50\text{N}$ ke beech mein hai — bilkul wahan jahan ek real pulling force honi chahiye. ✓

Phir se FBDs se build karte hain taaki dekh sakein kyun sirf $m_2$ ka weight system ko drive karta hai. $m_2$ ka FBD (latka hua, girta hai; positive = down): weight $m_{2}g$ neeche, tension $T$ upar. $m_{2}g-T=m_{2}a$ $m_1$ ka FBD (horizontally slide karta hai; positive = pulley ki taraf): sirf horizontal force $T$ hai (weight aur normal force vertically cancel ho jaate hain, dekho Normal Force). $T=m_{1}a$ Dono add karo $T$ cancel karne ke liye: $m_{2}g=(m_1+m_2)a\Rightarrow a=\frac{m_{2}g}{m_1+m_2}=\frac{6\times 10}{10}=\boxed{6{\text{m/s}}^2}$ Numerator sirf $m_{2}g$ hai — isliye hanging mass, table mass nahi, driving force set karta hai: sirf $m_2$ ki gravity string ke along kheenchti hai. Phir $T=m_{1}a$ se: $T=m_{1}a=4\times 6=\boxed{24\text{N}}$ Note karo $T=24\text{N}<m_{2}g=60\text{N}$: kyunki $m_2$ gir raha hai, string uska poora weight nahi thaam sakti.

Figure mein, daayein taraf ka red arrow applied force $F$ hai, aur notice karo ki yeh sirf front block ko touch karta hai — yeh ek visual fact poori solution ki key hai: back blocks sirf string tension feel karte hain, $F$ directly kabhi nahi. Labels $T_2$ (do rear blocks ke beech) aur $T_1$ (front pair ke beech) un do strings ko mark karte hain jo hume nikalni hain. Step 1 — Poore system ke liye $a$. Internal tensions poori chain mein cancel ho jaati hain, toh sirf $F$ total mass par act karta hai: $F=(m_1+m_2+m_3)a\Rightarrow a=\frac{12}{1+2+3}=\frac{12}{6}=\boxed{2{\text{m/s}}^2}$ Step 2 — $T_2$ ke liye back block $m_3$ ko isolate karo. $m_3$ par sirf horizontal force string tension $T_2$ hai jo ise aage kheenchti hai: $T_2=m_{3}a=3\times 2=\boxed{6\text{N}}$ Step 3 — $T_1$ ke liye $\{m_2,m_3\}$ ko saath isolate karo. Is pair par sirf bahari forward pull $T_1$ hai: $T_1=(m_2+m_3)a=5\times 2=\boxed{10\text{N}}$ Check: front block FBD deta hai $F-T_1=m_{1}a\Rightarrow 12-10=1\times 2=2$ ✓. Tensions peeche ki taraf chhoti hoti jaati hain kyunki har string sirf apne peeche ki mass ko accelerate karti hai.

FBD: weight $mg$ neeche, tension $T$ upar. Upar kheeencha gaya (positive = up): $T-mg=ma\Rightarrow T=m(g+a)=2(10+3)=\boxed{26\text{N}}$ Utara gaya, neeche accelerate karta hua (positive = down): $mg-T=ma\Rightarrow T=m(g-a)=2(10-3)=\boxed{14\text{N}}$ Same bucket, same string, do alag tensions — kyunki tension acceleration ke saath respond karta hai, sirf weight se nahi. Note karo $26>mg>14$: tez karne ke liye weight se zyada chahiye, dheere neeche karne ke liye kam.

Figure mein red block incline par $m_1$ hai; uska motion arrow slope ke upar point karta hai, aur hanging white $m_2$ ka arrow neeche hai — yahi direction hai jo hum ab numbers se justify karenge. Triangle ke base par marked angle $\theta =30^{\circ }$ wahi hai jo hum $\sin \theta$ mein feed karenge. Naya ingredient: incline par, gravity ka ek part slope ke along hota hai, $m_{1}g\sin \theta$, aur ek part slope ke andar press karta hai, $m_{1}g\cos \theta$. Sirf along-slope part tug-of-war mein compete karta hai; into-slope part Normal Force dwara cancel ho jaata hai. Hume $\sin \theta$ chahiye kyunki yahi gravity ka woh fraction hai jo motion ki direction ke along hota hai. $m_{1}g\sin \theta =4\times 10\times \frac{1}{2}=20\text{N},m_{2}g=30\text{N}$ Kyunki $m_{2}g=30>20$, hanging block jeet jaata hai aur girta hai; $m_1$ incline par upar slide karta hai (figure mein arrows se match karta hai). FBDs (positive = har block ke liye motion ki direction):

• Hanging $m_2$ (girta hai): $m_{2}g-T=m_{2}a$
• Incline $m_1$ (slope ke upar): $T-m_{1}g\sin \theta =m_{1}a$ Add karo $T$ cancel karne ke liye: $m_{2}g-m_{1}g\sin \theta =(m_1+m_2)a$ $a=\frac{(m_2-m_1\sin \theta )g}{m_1+m_2}=\frac{(3-4\cdot 0.5)\times 10}{7}=\frac{10}{7}\approx \boxed{1.43{\text{m/s}}^2}$ $T$ ke liye back-substitute karo: $T=m_2(g-a)=3\left(10-\frac{10}{7}\right)=3\cdot \frac{60}{7}=\frac{180}{7}\approx \boxed{25.7\text{N}}$ Check ranges: $20\text{N}<T\approx 25.7\text{N}<30\text{N}$ — tension do competing pulls ke beech mein hai, jaisa hona chahiye. ✓

Derived formula reuse karo, sign convention rakhte hue "positive = $m_2$ girta hai / $m_1$ up-slope": $a=\frac{(m_2-m_1\sin \theta )g}{m_1+m_2}=\frac{(1.5-4\cdot 0.5)\times 10}{4+1.5}=\frac{(1.5-2)\times 10}{5.5}=\frac{-5}{5.5}\approx \boxed{-0.91{\text{m/s}}^2}$ Negative sign physics bol raha hai: hamari assumed direction galat thi. Reality mein $m_1$ incline par neeche slide karta hai aur $m_2$ upar uthta hai, magnitude $|a|\approx 0.91{\text{m/s}}^2$ ke saath. Tension. Tension string ke along ek pulling force hai, aur uski strength (newtons ki quantity) ek single value hai jo is baat par depend nahi karta ki humne kaun sa direction positive choose kiya — sign convention flip karne se $a$ ka sign flip hoga lekin actual pull unchanged rahega. Toh hum signed value $a\approx -0.91{\text{m/s}}^2$ seedha $m_2$ ki equation mein substitute kar sakte hain aur strength padh sakte hain: $T=m_2(g-a)=1.5\left(10-(-0.91)\right)=1.5\times 10.91\approx \boxed{16.4\text{N}}$ Isliye hum algebraic sign rakhte hain direction guess karne ki jagah: equations khud sachchi baat bata dete hain.

Equal masses ⇒ $a=\frac{(m_1-m_2)g}{m_1+m_2}=0$, toh system static hai: $T=m_{1}g=5\times 10=\boxed{50\text{N}}\text{dono sides par}.$ Equal kyun: ek massless string element $T_2-T_1=(dm)a=0$ follow karta hai, toh tension point-to-point change nahi ho sakta — frictionless, massless pulley ke across bhi nahi, jiske paas tension difference banana ke liye koi rotational inertia nahi hai. Dekho Frictionless Pulleys vs Pulleys with Inertia.

$a=\frac{(m_1-m_2)g}{m_1+m_2}$ aur $T=\frac{2m_1{m}_2}{m_1+m_2}g$ se shuru karo.

(a) $m_1=m_2=m$: equal masses substitute karo: $a=\frac{(m-m)g}{m+m}=\frac{0}{2m}g=\boxed{0},T=\frac{2m\cdot m}{m+m}g=\frac{2m^2}{2m}g=\boxed{mg}.$ Balanced — har side sirf apna weight pakad rahi hai, koi motion nahi. ✓

(b) $m_2=0$: doosre mass ki jagah zero substitute karo: $a=\frac{(m_1-0)g}{m_1+0}=\frac{m_{1}g}{m_1}=\boxed{g},T=\frac{2m_1\cdot 0}{m_1+0}g=\boxed{0}.$ Doosri taraf kuch nahi ⇒ bacha hua mass free fall mein hai, aur koi mass nahi kheenchne ke liye string slack ho jaati hai (zero pull). ✓

(c) $m_1\to \infty$ ($m_2$ fixed ke saath): har formula ke top aur bottom ko $m_1$ se divide karo, phir $m_2/m_1\to 0$ hone do: $a=\frac{(1-m_2/m_1)g}{1+m_2/m_1}\to \frac{(1-0)g}{1+0}=\boxed{g},T=\frac{2m_2}{1+m_2/m_1}g\to \frac{2m_2}{1+0}g=\boxed{2m_{2}g}.$ Limitlessly bhaari $m_1$ $g$ par girta hai, aur halke mass ko itna zyada upar kheenchta hai ki string tension $2m_{2}g$ ke paas pahunch jaati hai — yeh halke mass ka weight $m_{2}g$ plus extra force $m_{2}g$ hai jo ise $g$ par upar accelerate karne ke liye chahiye. ✓

Static single hanger, $T=?$ ::: $T=mg$ (sirf isliye kyunki $a=0$). Ek inextensible string par bandhi masses mein kaun si quantity share hoti hai? ::: acceleration magnitude $a$. $F$ se kheenchi gayi blocks ki chain: kaunsi string mein sabse zyada tension hai? ::: pull ke sabse paas wali (woh apne peeche sabse zyada mass accelerate karti hai). Bucket $a$ par upar accelerate kiya, tension? ::: $T=m(g+a)$, $mg$ se zyada. Bucket $a$ par utara gaya, tension? ::: $T=m(g-a)$, $mg$ se kam. Incline par, gravity ka kitna fraction motion drive karta hai? ::: along-slope part $mg\sin \theta$. Negative computed $a$ ka matlab? ::: sach muchi ki motion tumhari assumed positive direction ke opposite hai. Atwood limit $m_1\to \infty$: $a$ aur $T$? ::: $a\to g$, $T\to 2m_{2}g$.